    
\(\mathcal {M}\)  is  not orthomodular. For an example let \(L:=h(< 0)\cap k(> 2)\)  and \(M:=h(< 0)\cap k(> 0)\) . Then clearly \(L\subset M\) . By (REF )(b), \(L,M \in \mathcal {M}\)  with \(L^\perp =h(\ge 0)\cap k(\le 2)\) , \(M^\perp =h(\ge 0)\cap k(\le 0)\) , and \( L^\perp \cap M = h(\ge 0)\cap k(\le 2)\cap h(< 0)\cap k(> 0)=\emptyset \) . Hence \((L\cup M^\perp )^\wedge =(L^\perp \cap M)^\perp =\mathbb {R}^4\) . Therefore \((L\vee M^\perp )\wedge M=(L\cup M^\perp )^\wedge \cap M=M\ne L\) ,
and orthomodularity does not hold.The presumed proof of orthomodularity of \(\mathcal {M}\)  in [1]} is not correct, since there one infers \(\mathfrak {x}\in (L\cup M^\perp )^\sim \cap M\)  from \(\mathfrak {x}\in (L\cup M^\perp )^\wedge \cap M\) . Indeed, this does not hold true in general. In the above example one has \((L\cup M^\perp )^\wedge \cap M=M\)  and \((L\cup M^\perp )^\sim \cap M=L\)  with \(L\subsetneq M\)  by (REF ) as \(L=\hat{\Gamma },M= \Delta ^\perp \) .
